Overview
This lecture covers how to solve implicit differentiation problems, including finding first and second derivatives, applying the product and chain rules, and evaluating derivatives at specific points.
Implicit Differentiation Basics
- When differentiating y-terms with respect to x, multiply by dy/dx since y is a function of x.
- For example: d/dx(y³) = 3y² dy/dx and d/dx(r⁴) = 4r³ dr/dx.
Problem 1: Circle Equation
- Original equation: x² + y² = 100.
- Differentiate both sides: 2x + 2y dy/dx = 0.
- Isolate dy/dx: dy/dx = -x/y.
- At point (6, 8): dy/dx = -3/4.
Problem 2: Cubic Equation with Product Rule
- Original equation: x³ + 4xy + y² = 13.
- Differentiate: 3x² + 4y + 4x dy/dx + 2y dy/dx = 0.
- Factor dy/dx: (4x + 2y) dy/dx = -3x² - 4y.
- Solve for dy/dx: dy/dx = (-3x² - 4y)/(4x + 2y).
- At point (1, 2): dy/dx = -11/8.
Problem 3: Implicit Differentiation with Trig Functions
- Equation: 5 - x² = sin(xy²).
- Differentiate: -2x = cos(xy²) [y² + 2xy dy/dx].
- Rearranged: -2x - y² cos(xy²) = 2x y cos(xy²) dy/dx.
- Solve for dy/dx: dy/dx = [-2x - y² cos(xy²)] / [2x y cos(xy²)].
Problem 4: Second Derivative by Implicit Differentiation
- Equation: x³ + y³ = 9.
- First derivative: dy/dx = -x²/y².
- Use quotient rule for d²y/dx².
- After simplification: d²y/dx² = [-2x y³ - 2x⁴] / y⁵ or -2x(y³ + x³)/y⁵.
- At point (1, 2): d²y/dx² = -9/16.
Key Terms & Definitions
- Implicit Differentiation — Differentiating equations involving x and y, treating y as a function of x and multiplying its derivatives by dy/dx.
- Product Rule — d/dx [fg] = f' g + f g'.
- Chain Rule — Differentiating a composed function: d/dx[f(g(x))] = f'(g(x)) g'(x).
- Quotient Rule — d/dx[f/g] = [g f' - f g']/g².
- Second Derivative — The derivative of the first derivative, representing the rate of change of the slope.
Action Items / Next Steps
- Practice similar implicit differentiation problems, including those with trigonometric and higher-order derivatives.
- Review product, chain, and quotient rules for upcoming assignments.